Fields of force
Worked examples
1.
An object in a field is subject to a force F from the field. How much work is required to move
the object a distance d at right angles to the field lines? What is the change in potential of the
object?
2.
Sketch the equipotential lines above the surface of the Earth.
3.
Sketch the equipotential lines created by the following field pattern:
Solutions
1.
The work done is
W  ( F cos )  d  ( F cos 90 )  d  0
If no work is done, no energy is supplied to the object, and there is no change in potential!
Equipotential lines are always at right angles to field lines.
2.
3.
Gravitational fields
Worked examples
1.
A satellite with a mass of 50 kg is orbiting the Earth at a height of 35900 km above the
surface. What is the gravitational field strength at this height? What is the force on the
satellite? (The Earth’s mass and radius are 5.98 × 10 24 kg and 6380 km.)
2.
The Sun has a mass of 1.99 × 1030 kg and a radius of 6.96 × 108 m. What would a 50 kg
person weigh on the surface of the Sun?
3.
What is the gravitational force between two cars, each weighing 5000 kg, a distance 5 m
apart?
Solutions
1.
The radius of the orbit is 35900 km  6380 km  42280 km .
At this distance, the gravitational field strength is
GM E 6.67 1011  5.98 1024
g 2 
 0.223 N kg -1
7 2
r
(4.228 10 )
The force on the satellite is
F  mg  50  0.223  11.15 N
2.
The person’s weight is simply the gravitational force. It is
GM S m 6.67 10 11 1.99 1030  50
F

 13700 N
rS2
(6.96 108 ) 2
This is a lot! Note that standing on the surface of the Sun is impossible – it’s a gas, not a solid,
besides being very hot.
3.
The force is F

Gm2 6.67 1011  (5 103 ) 2

 6.67 105 N . Note how small this
d2
52
is! The gravitational force is very weak, which makes it hard to measure unless you have a
large amount of mass.
Circular motion in the solar system
Worked examples
1.
At what distance above the Earth’s surface must a communications satellite be placed to be in
geostationary orbit?
2.
Calculate the kinetic and gravitational potential energy of a 50 kg satellite orbiting the Earth at
a height of 3000 km.
3.
A star five times as massive as the Sun collapses to a radius of 5 km after burning all its fuel.
What is the escape velocity from the surface of the star?
Solutions
1.
The orbital period, T, must be exactly one day, or 8.64 × 10 4 s. This means that the angular
velocity is

2
2

 7.27 10 5 rad s -1
T 8.64 10 4
We now equate the centripetal force with the gravitational force to find the orbital radius:
GmM E
r2
GM E
r3 
 7.56 10 22
2
m2 r 

r  4.23 107 m
Finally, we must subtract the radius of the Earth to find the height above the surface:
h  r  rE  4.23 107  6.3 106  3.59 107 m  35900 km
2.
The orbital radius is
r  rE  h  6.3 106  3.0 106  9.3 109 m .
The kinetic energy is calculated by equating the centripetal force to the gravitational force:
mv2 GM E m

r
r2
1 2 GM E m 6.67 10 11  5.98 10 24  50
mv 

 1.07 109 J
6
2
2r
2  9.3 10
The potential energy is given by the formula:

3.
GM E m
6.67 10 11  5.98 10 24  50

 2.14 109 J
6
r
9.3 10
The mass of the Sun is 1.99 × 1030 kg. This means that the star has mass 5 × 1.99 × 1030 kg =
9.95 ×1030 kg. The escape velocity is therefore
vE 
2GM
2  6.67 10 11  9.95 1030

 3.64 108 m s -1
3
R
5 10
This is greater than the speed of light (3.0 × 10 8 m s-1), the fastest speed possible! Nothing can
escape from the surface of the star, not even light – it is a black hole.
Electric fields
Worked examples
1.
What is magnitude of the electric field at a distance of 0.1 μm from an electron (charge 1.6 ×
10-19 C)?
2.
What is the electric force between two electrons separated by a distance of 0.1 μm?
3.
Two parallel metal plates of area 20 cm2 are charged with ±1.6 × 10-12 C. What is the electric
field between the plates?
Solutions
1.
We need the electric field due to a point charge. The formula is
E
2.
Q
40 d 2

1.6 1019
 1.44 105 N C -1
12
6 2
4  8.85 10  (0.110 )
To solve this, we could simply use Coulomb’s law. However, in the previous question, we
have already calculated the field at a distance of 10 μm from an electron. We are now placing
another electron at this point. The force is then
F  EQ  14.4 1.6 1019  2.3 1012 N
We can compare this with the result we get from Coulomb’s law:
F
3.
QQ
(1.6 10 19 ) 2

 2.3 10 12 N
40 d 2 4  8.85 10 12  (0.1106 ) 2
Remembering to use the correct units, the charge per unit area on each plate is
1.6 10 12

 8 10 10 C m -2
4
20 10
One plate is positive, the other is negative:
The field between the plates is

8 10 10
E 
 90.4 N C -1
12
 0 8.85 10
Electric potential
Worked examples
1.
What is the electric potential due to a charge of -4.5 × 10-10 C at a distance of 80 cm?
2.
Two large, parallel metal plates are separated by a distance of 20 mm. The potential difference
between the plates is 5 V. What is the electric field in the gap between the plates? The area of
each plate is 20 cm2. What is the charge on each plate?
3.
What is the electric potential at a distance of 1.5 × 10 -9 m from a proton? (Proton charge = 1.6
× 10-19 C.) What is the potential energy when an electron (charge -1.6 × 10-19 C) is placed at
this distance?
Solutions
1.
The potential due to a point charge is
V
2.
Q
40 d

 4.5 10 10
 5.06 V
4  8.85 10 12  0.8
The electric field between the plates in uniform, with a value equal to the potential gradient:
E
V
5

 250 V m -1
3
x
20 10
We can use our conducting-sheet formula to work out the charge per unit area on the plates:
E

0
   0 E  8.85 10 12  250  2.2110 9 C m -2
Finally, knowing the area and the charge per unit area allows us to calculate the charge:
Q  A  2.21109  20 104  4.42 1012 C
3.
The potential due to the proton (a point charge) is
1.6 1019
V

 0.96 V
40 d 4  8.85 10 12 1.5 109
Qp
When we place an electron at this position, the potential energy is
QeV  1.6 10 19  0.96  1.54 10 19 J
Note the difference between electric potential and potential energy. Also note how small this
energy is! We will describe a more appropriate unit of energy for these situations (the
electronvolt) in a later section.
Magnetic fields
Worked examples
1.
A wire of length 20 cm carrying a current of 300 mA is placed in a uniform magnetic field.
When the wire is perpendicular to the field, the force on the wire is measured to be 0.024 N.
What is the strength of the field? What would be the force if the wire were at an angle of 30°
to the field?
2.
The coil of an electric motor has dimensions 20 mm × 2 mm, contains 50 turns and carries a
current of 30 mA. What is the maximum torque when the coil is subjected to a field of 0.1 T?
What is the torque when the coil makes angle of 60° with the field?
3.
A proton travelling at 500 km s-1 is subjected to a magnetic field perpendicular to the direction
of motion. What field strength is required to make the proton travel in a circle of radius 3 m?
(The proton charge is 1.6 × 10-19 C, the mass is 1.67 × 10-27 kg.)
Solutions
1.
The field strength is given by
B
F
0.024

 0.4 T
IL 0.2  0.3
When the wire is at angle of 30° to the field, the force is reduced.
F  BIL sin   0.4  0.3 0.2  sin 30  0.012 N
2.
The diagram shows the arrangement at the point of maximum torque:
Torque  Fd  NBILd  50  0.1 0.01 0.03  0.002  3 106 N m .
When the coil makes an angle of 60° with the field, the force is the same, but the torque is
reduced by a factor of sin 30°, as can be seen from the diagram:
The new torque is 3 × 10-6 × sin 30° = 1.5 × 10-6 N m.
3.
The radius of the arc is given by the formula
r
mv
BQ
We may rearrange this to find the field strength:
B
mv 1.67 1027  500 103

 1.74 mT
rQ
3 1.6 1019
Applications of electric and magnetic fields
Worked examples
1.
What is the velocity of an electron given a kinetic energy of 1.4 keV? (The electron mass is
9.1 × 10-31 kg.)
2.
A velocity selector for electrons uses a B-field of 0.7 T and an E-field of 5.6 × 105 V m-1.
What is the kinetic energy of the emerging electrons (in eV)?
3.
What is the cyclotron frequency for protons travelling in a magnetic field of strength 0.7 T?
(The proton mass is 1.67 × 10-27 kg, the charge is 1.6 × 10-19 C.) If the maximum radius is 1.2
m, what is the maximum energy the proton could reach?
Solutions
1.
First, we convert the energy into joules:
1.4 keV  1.4 103 1.6 10 19 J  2.24 10 16 J
We then equate this to the kinetic energy to find the velocity.
1 2
mv  2.24 10 16
2
2  2.24 10 16
v
 2.22 107 m s -1
31
9.110
This is very fast (almost 1/10 of the speed of light)!
2.
The selected velocity is
v
E 5.6 105

 8 105 m s -1
B
0.7
Electrons at this velocity have kinetic energy
1 2 1
mv   9.110 31  (8 105 ) 2  2.9110 19 J
2
2
Finally, we convert this energy into electronvolts:
2.9110
3.
19
2.911019
J
eV  1.82 eV
1.6 1019
The cyclotron frequency is
f 
QB
1.6 10 19  0.7

 1.07 107 Hz
 27
2m 2 1.67 10
The velocity of the proton when the radius is 1.2 m is
v  r  2fr  2 1.07 107 1.2  8.07 107 m s -1
At this velocity, the proton’s kinetic energy is
1 2 1
mv  1.67 10  27  (8.07 107 ) 2  5.44 10 12 J  34 MeV
2
2
Electromagnetic induction
Worked examples
1.
A straight wire of length L is travelling at velocity v m s-1 in a direction perpendicular to a
magnetic field of strength B. What is the EMF generated between the ends of the wire? Apply
your result to find the EMF generated between the wing-tips of an aircraft (48 m apart) flying
at 640 km h-1 due to the Earth’s magnetic field (whose vertical component is 0.6 × 10 -4 T).
2.
A coil of resistance 3.5 Ω has a steady current of 0.2 A flowing through it. The coil is then
shorted, and the current decay is measured. After 0.12 s the current has fallen to 0.1 A. What
is the inductance of the coil?
3.
An electronic component has a resistance of 25 Ω and an inductance of 3 mH. If the potential
difference across the component is suddenly increased from 0 to 2 V, how long does it take
for the current to reach 0.03 A? If this operation constitutes one cycle, how many cycles per
second is the component capable of?
Solutions
1.
The area swept out by the wire in one second is A = Lvt m2. The EMF is given by the rate of
change of flux:
d
dt
dA
 B
since B is constant 1.67 × 10-27 kg
dt
  BLv
E
Applying this formula to the aircraft (and converting the speed into m s -1) gives an EMF of
E   0.6 104  48  640 103 3600  0.512 V
2.
After the coil is shorted out, the current decays at a rate e
is the inductance of the coil. This means that
 Rt L
, where R is the resistance and L
0.1  0.2e 3.50.12 L
 3.5  0.12
0.1
 log
L
0.2
0.42
L
 606 mH
log 2
3.
The steady-state current which would eventually be obtained is
I
V
2

 0.04 A
R 25
The current after time t is I  I 0 (1  e
terms of the current:
 Rt | L
) . Rearranging this equation gives the time in
t
L
I
log( 1  )
R
I0
0.003
0.03
log( 1 
)
25
0.04
 0.166 ms

If this is the time taken for one cycle, the maximum operating frequency is
1
 6.02 kHz
0.166 10 3